Thanks for the rapid response!  I think that I have a mental block regarding 
this issue.  Perhaps I am looking at the fusion process from the wrong angle.

Help me understand where my error is as I do not follow your example below.  
Let's try a mental experiment.  I can see that it would be possible to build a 
machine that accelerates protons until they reach the required threshold level 
of 5.6 MeV.  It would take at least this much energy to generate the projectile 
proton.  Now, when the bullet proton comes close to the nucleus the fusion 
takes place.  At this point in time it has lost all of its kinetic energy and 
is absorbed into the nucleus.  I assume that very soon thereafter the 3.41 MeV 
is released as a gamma or some other form of radiation.  The tables of nuclides 
informs me that the copper 59 has a half life of 81.5 seconds before the beta 
plus decay so the energy associated with that process must be in place but does 
not exit until later.

Could you help me understand how the 5.6 MeV is recovered or released?   Is 
there extra energy released into the copper crystal structure that equals this 
magnitude?   I am having a difficult time trying to get back the 5.6 MeV to 
make the next proton energetic enough for the next reaction.  Forgive me for 
being ignorant about this mechanism, but it truly is difficult to visualize.

Dave 



-----Original Message-----
From: Finlay MacNab <finlaymac...@hotmail.com>
To: vortex-l <vortex-l@eskimo.com>
Sent: Tue, May 22, 2012 1:50 pm
Subject: RE: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?



Your calculation does not take into account the fact that the activation energy 
barrier releases the energy added to overcome it during the reaction.  In this 
case once coulomb repulsion is overcome, the energy is added back to the system 
by attractive nuclear force.  The 3.41MeV is the change in mass energy balance 
after the reaction, what happens in between is not important to the calculation.


Analogously, If a person descends from the 3rd to second floor of a building, 
they are just as close to the ground as if they climb from the 3rd to the 53rd 
floor before climbing back down the the second floor to end their journey.  


This is the nature of all "activation energy barriers".





To: vortex-l@eskimo.com
Subject: [Vo]: Proton Fusion Ni58 to Cu59 Endothermic?
From: dlrober...@aol.com
Date: Tue, 22 May 2012 13:35:41 -0400


I have been researching the cold fusion reaction that is suggested by Rossi and 
Focardi in their recent paper  
http://www.journal-of-nuclear-physics.com/files/Rossi-Focardi_paper.pdf and 
have a couple of questions.  The authors suggest that 3.41 MeV of energy is 
released by the fusion of a proton with a nickel 58 nucleus into copper 59.  I 
can obtain this value if I calculate the mass difference between a copper 59 
atom and a nickel 58 atom plus the mass of the proton and the mass of the extra 
electron.
So far their calculations are in line with mine.  The problem arises when I 
consider the amount of energy required to overcome the coulomb barrier in order 
to activate the fusion.  The two authors seem to overlook this entirely when 
they calculate the energy available from their proposed reaction.  The chart on 
page 5 of their paper shows that 3.41 MeV is released at the conclusion to the 
reaction but no allowance is given to the energy needed to initiate it.
They do mention the activation energy in their theoretical interpretation on 
page 7.  In this section they calculate that it takes 5.6 MeV to overcome the 
barrier.  The authors use assumed values for the closeness required and thus 
energy barrier in their example.
With these two numbers available I make the assumption that there is a net 
energy requirement of 5.6 MeV – 3.41 MeV or 2.19 MeV for the fusion.  Is there 
a reason that my calculation is in error?  Does the 3.41 MeV have hidden within 
it the activation energy?  I can see no good reason to suspect that this is the 
case since it would be possible for a device to send high speed protons into a 
target made of copper.
The copper would then shed the 3.41 MeV by some means and that would obviously 
not repay the debt.  Of course I understand that the following beta plus decay 
would release an additional significant amount of energy as the copper 
transforms into nickel 59.  I calculate this energy as 5.8 MeV when the 
released positron is annihilated.  This value matches that of the two authors 
which I assume is correct.
A recap of the question is: Is the fusion of nickel 58 with a proton and 
electron into copper 59 an endothermic reaction?
Dave


Reply via email to